Question: s involving definite integrals (algebraic) AP.CALC: CHA‑4 (EU), CHA‑4.D (LO), CHA‑4.D.1 (EK), CHA‑4.D.2 (EK), CHA‑4.E (LO), CHA‑4.E.1 (EK) Google Classroom Facebook Twitter Email You might need: Calculator Problem The average cost per meal served at Kiran's restaurant decreases at a rate of $\dfrac{2400}{q^2}$ dollars per meal served that month (where $q$ is the number of meals served). By how many dollars does the average cost per meal decrease between $q=300$ and $q=360$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $0.67$ (Choice B) B $0.81$ (Choice C) C $1.11$ (Choice D) D $1.33$
Answer: Letting $r(q)$ be the decrease in the average cost when Kiran's restaurant serves $q$ meals compared to serving $1$ meal, we are given that $r'(q)=\dfrac{2400}{q^2}$. We want to find $r(360)-r(300)$. According to the Fundamental Theorem of Calculus, $\begin{aligned} r(360)-r(300)&=\int_{300}^{360} r'(q)\,dq \\\\ &=\int_{300}^{360}\left(\dfrac{2400}{q^2}\right)dq \end{aligned}$ $\int_{300}^{360}\left(\dfrac{2400}{q^2}\right)dq = \dfrac{4}{3}\approx1.33$ In conclusion, between $q=300$ and $q=360$ the average cost per meal decreases by $\$1.33$.